Let \(A\) be a compact subspace of a Hausdorff space \(X\). Show that \(A\) is closed.
Solution
1. It suffices to show \(X \setminus A\) is open.
a set is closed iff its complement is open.
2. Let \(x \in X \setminus A\).
take an arbitrary point outside \(A\).
3. For each \(a \in A\), there are disjoint open sets \(U_a \ni x\) and \(V_a \ni a\).
\(X\) is Hausdorff and \(x \neq a\).
4. \(\{V_a\}_{a \in A}\) is an open cover of \(A\).
each \(a \in A\) lies in \(V_a\).
5. Since \(A\) is compact, there is a finite subcover \(A \subseteq V_{a_1} \cup \cdots \cup V_{a_k}\).
compactness.
6. Let \(U = U_{a_1} \cap \cdots \cap U_{a_k}\).
define a neighborhood of \(x\).
7. \(U\) is open, contains \(x\), and is disjoint from \(A\).
\(U\) is a finite intersection of open sets containing \(x\); and \(U \cap V_{a_i} = \varnothing\) for each \(i\) (since \(U \subseteq U_{a_i}\) and \(U_{a_i} \cap V_{a_i} = \varnothing\)), so \(U \cap A = \varnothing\) by 5.
8. Hence \(x\) is an interior point of \(X \setminus A\), so \(X \setminus A\) is open.
7 holds for every \(x \in X \setminus A\).
9. Therefore \(A\) is closed.
1 and 8.
10. Q.E.D.
9.